GATE CE 2015 SET-2

Question 1
While minimizing the function f(x), necessary and sufficient conditions for a point, x_{0} to be a minima are:
A
{f}'(x_{0}) \gt 0\: and\: \: {f}''(x_{0})=0
B
{f}'(x_{0}) \lt 0\: and\: \: {f}''(x_{0})=0
C
{f}'(x_{0})=0\: and\: \: {f}''(x_{0}) \lt 0
D
{f}'(x_{0})=0\: and\: \: {f}''(x_{0}) \gt 0
Engineering Mathematics   Calculus
Question 1 Explanation: 
f(x) has a local minimum at x=x_{0}
if {f}'\left ( x_{0} \right )=0
and {f}''\left ( x_{0} \right ) \gt 0
Question 2
In Newton-Raphson iterative method, the initial guess value ( x_{ini}) is considered as zero while finding the roots of the euation:f(x)=-2+6x-4x^{2}+0.5x^{3}. The correction, \Delta x, to be added to x_{ini} in the first iteration is____________.
A
0.5
B
0.33
C
0.8
D
0.2
Engineering Mathematics   Numerical Methods
Question 2 Explanation: 
\begin{aligned} f\left ( x \right )&=-2+6x-4x^{2}+0.5x^{3} \\ {f}'\left ( x \right )&=6-8x+1.5x^{2} \\ x_{ini}&=0\\ \text{By Newton }&\text{Raphson Method,}\\ x_{1}&=x_{ini}-\frac{f\left ( x_{ini} \right )}{{f}'\left ( x_{ini} \right )} \\ &= 0-\frac{-2}{6} \\ \Rightarrow \;\;x_{1}&=\frac{1}{3} \\ \therefore \;\; \Delta x&=x_{1}-x_{ini}=\frac{1}{3} \end{aligned}
Question 3
Given , i=\sqrt{-1}, the value of the definite integral, I=\int_{0}^{\pi/2}\frac{\cos x+i\sin x}{\cos x-i\sin x}dx is:
A
1
B
-1
C
i
D
-i
Engineering Mathematics   Calculus
Question 3 Explanation: 
\begin{aligned} I&=\int_{0}^{\pi /2}\frac{\cos x+i\sin x}{\cos x-i\sin x}dx \\ &=\int_{0}^{\pi /2}\frac{e^{ix}}{e^{-ix}}dx=e^{2ix}dx \\ &=\frac{1}{2i}\left [ e^{2ix} \right ]_{0}^{\pi /2} \\ &=\frac{1}{2i}\left [ e^{i\pi }-1 \right ] \\ &=\frac{1}{2i}\left ( -1-1 \right ) \\ &=-\frac{1}{i}=i \end{aligned}
Question 4
\lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{2x} is equal to
A
e^{-2}
B
e
C
1
D
e^{2}
Engineering Mathematics   Calculus
Question 4 Explanation: 
y=\: \lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{2x}
\log y = \lim_{x\rightarrow \infty }2x\log \left ( 1+\frac{1}{x} \right )
Which is in the form of \infty \times 0.
To convert this into \frac{0}{0} form, we rewrite as,
\Rightarrow \log y= \lim_{x\rightarrow \infty }\frac{2\log \left ( 1+\frac{1}{x} \right )}{1/x}
Now is in \frac{0}{0} form.
Using L'Hospital's Rule,
\begin{aligned} \log y&=\lim_{x\rightarrow \infty }\frac{\frac{2\times -\frac{1}{x^{2}}}{1+\frac{1}{x}}}{-\frac{1}{x^{2}}} \\ \log y&=\lim_{x\rightarrow \infty }\frac{2}{1+\frac{1}{x}}=2 \\ \therefore \;\; y&= e^{2}\end{aligned}
Question 5
Let \mathbf{A}=\left [ a_{ij} \right ],\; \; \; 1\leq i, \; j\leq n with n\geq 3\; and\; a_{ij}=i.j. The rank of A is:
A
0
B
1
C
n-1
D
n
Engineering Mathematics   Linear Algebra
Question 5 Explanation: 
Rank of A=1
Because each row will be scalar multiple of first row.So we will get only one-zero row in row Echeleaon form of A.

Alternative:
Rank of A=1
Because all the minors of order greater than 1 will be zero.
Question 6
A horizontal beam ABC is loaded as shown in the figure below. The distance of the point of contraflexure from end A (in m) is _________.
A
1.25
B
0.25
C
0.5
D
0.75
Solid Mechanics   Shear Force and Bending Moment
Question 6 Explanation: 
Reaction at B

\begin{aligned} \Delta_{\mathrm{B}}&=0 \text { (Compatibility condition) }\\ \frac{10 \times(0.75)^{3}}{3 E I}&+\frac{2.5 \times 0.75^{2}}{2 E I}-\frac{R_{B} \times 0.75^{3}}{3 E I}=0 \\ \frac{9 R_{B}}{64 E I}&=\frac{135}{64 E I}\\ \therefore \quad \mathrm{R}_{\mathrm{B}}&=15 \mathrm{kN} \end{aligned}
BM at a distance x from free end
Reaction at B

\begin{aligned} & & \mathrm{BM}_{x} &=10 \times x-15 \times(x-0.25)=0 \\ \Rightarrow & & 10 x &=15 x-3.75 \\ \Rightarrow & & 5 x &=3.75 \\ \therefore & & x &=0.75 \mathrm{m} \end{aligned}
\therefore From end A, distance is 0.25m
Question 7
For the plane stress situation shown in the figure, the maximum shear stress and the plane on which it acts are:
A
-50 MPa, on a plane 45^{\circ} clockwise w.r.t. x-axis
B
-50 MPa, on a plane 45^{\circ} anti-clockwise w.r.t. x-axis
C
50 MPa, at all orientations
D
Zero, at all orientations
Solid Mechanics   Principal Stress and Principal Strain
Question 7 Explanation: 
Under hydrostatic loading condition, stresses at a point in all directions are equal and hence no shear stress.
Alternatively,
\tau=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{50-50}{2}=0
Thus, Mohr's circle reduces to a point.
Hence shear stress at all orientations is zero.
Question 8
A guided support as shown in the figure below is represented by three springs (horizontal, vertical and rotational) with stiffness k_{x},k_{y}\: and \: k_{\theta } respectively. The limiting values of k_{x},k_{y}\: and \: k_{\theta } are:
A
\infty ,0,\infty
B
\infty ,\infty ,\infty
C
0 ,\infty ,\infty
D
\infty ,\infty ,0
Structural Analysis   Determinacy and Indeterminacy
Question 9
A column of size 450 mm x 600 mm has unsupported length of 3.0 m and is braced against side sway in both directions. According to IS 456:2000, the minimum eccentricities (in mm) with respect to major and minor principle axes are:
A
20.0 and 20.0
B
26.0 and 21.0
C
26.0 and 20.0
D
21.0 and 15.0
RCC Structures   Footing, Columns, Beams and Slabs
Question 9 Explanation: 


e_{\min }=\operatorname{maximum}\left\{\begin{array}{l} \frac{L}{500}+\frac{D}{30} \\ 20 \mathrm{mm} \end{array}\right.
x-x will be major axis and y-y will be minor axis.
\begin{aligned} \therefore \quad e_{\min y y}&=\frac{3000}{500}+\frac{450}{30}=21 \mathrm{mm} \\ \text{and }\quad e_{\min xx}&=\frac{3000}{500}+\frac{600}{30}=26 \mathrm{mm} \end{aligned}
Question 10
Prying forces are:
A
shearing forces on the bolts because of the joints
B
tensile forces due to the flexibility of connected parts
C
bending forces on the bolts because of the joints
D
forces due the friction between connected parts
Design of Steel Structures   Structural Fasteners
There are 10 questions to complete.

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